WWEKANE
Badges: 11
Rep:
?
#1
Report Thread starter 1 year ago
#1
I've been stuck on this since the start of February. I still can't get around the fact that in standard enthalpy of combustion and formation why hess cycles can be used.


I know that carbon combusts and reacts with oxygen to form co2 and hydrogen with water. I know benzene also combusts to form both co2 and water. But why do we say that the energy required to combust benzene is equal to the enthalpy of carbon dioxide and water reacting to form benzene when chemically this wouldn't happen. I also don't know why we put co2 and water for combustion products because even though carbon combusts to form co2 it doesn't water so why do we add this. Anyone help much appreciated

Thanks
0
reply
Kian Stevens
Badges: 16
Rep:
?
#2
Report 1 year ago
#2
You have to consider the first law of thermodynamics, i.e. the conservation of energy.
You're right; chemically speaking, you couldn't just produce benzene simply by reacting CO2 and H2O.
So, you're correct in asking why we must assume that the enthalpy change of combustion of benzene is in equal magnitude to the enthalpy change of reaction of CO2 + H2O... It all comes back to the first law of thermodynamics.

Let's say that the energy released by a system, as it undergoes the process A \rightarrow B, is -x kJ mol-1.
This energy is released in the form of electromagnetic radiation, and is due to the electrons in A lowering in energy levels to form stronger, more favourable bonds in B.
However, this lost energy must be conserved. The energy the electrons lose whilst going through the process A \rightarrow B must be equal to the energy they gain again, in order to increase in energy levels and undergo the reverse process B \rightarrow A. Therefore, the energy gained by the system whilst going through the process of B \rightarrow A would be +x kJ mol-1.

Since enthalpy change of combustion values are so exothermic, this means that these process are extremely favourable, as the bonds formed in the products are much more stable.
However, this also means that the reverse reaction (i.e. to go from CO2 and H2O back to whatever you combusted) is equally endothermic. Because of this, it's extremely unfavourable to have CO2 and H2O forming a relatively much less stable molecule, hence why you wouldn't observe them reacting to form things like benzene.

As for your last question, elemental carbon would oxidise to only form either carbon monoxide or carbon dioxide. There's no hydrogen present to contribute to the formation of any water.
Last edited by Kian Stevens; 1 year ago
1
reply
WWEKANE
Badges: 11
Rep:
?
#3
Report Thread starter 1 year ago
#3
This is fantastic. I haven't studied the movement of electrons and now this causes release in energy. But this explanation is fantastic. Thank you so much.

Would a model of potential energy. Ie the electrons further from nucleus have more potential energy be a good way to explain this for basic understanding. Much appreciate your help
0
reply
adviceiguess
Badges: 6
Rep:
?
#4
Report 1 year ago
#4
(Original post by WWEKANE)
This is fantastic. I haven't studied the movement of electrons and now this causes release in energy. But this explanation is fantastic. Thank you so much.

Would a model of potential energy. Ie the electrons further from nucleus have more potential energy be a good way to explain this for basic understanding. Much appreciate your help
For a basic understanding, yes that would be fine.
1
reply
Kian Stevens
Badges: 16
Rep:
?
#5
Report 1 year ago
#5
(Original post by WWEKANE)
This is fantastic. I haven't studied the movement of electrons and now this causes release in energy. But this explanation is fantastic. Thank you so much.

Would a model of potential energy. Ie the electrons further from nucleus have more potential energy be a good way to explain this for basic understanding. Much appreciate your help
Yes, this is the energy they lose.

I'm not sure if you've studied orbitals, and you probably won't have looked at hybridisation, but when an atom bonds with another, their atomic orbitals undergo something called hybridisation.
When this happens, the orbitals form a set of new orbitals called hybrid atomic orbitals, and these are said to be degenerate, i.e. they all have the same energy.
For this to happen, the initial atomic orbitals must lose energy, i.e. lower in energy levels; or as you said, lower in potential energy. This energy lost is what's released as electromagnetic radiation. These hybrid atomic orbitals overlap to form bonds.
Of course, as I said in my previous reply, the reverse is true; an equal amount of energy can be gained by the bond to break it, and so reverse the entire process, all due to the first law of thermodynamics (i.e. conservation of energy).

Glad I could help!
Last edited by Kian Stevens; 1 year ago
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Have you experienced financial difficulties as a student due to Covid-19?

Yes, I have really struggled financially (36)
17.06%
I have experienced some financial difficulties (62)
29.38%
I haven't experienced any financial difficulties and things have stayed the same (82)
38.86%
I have had better financial opportunities as a result of the pandemic (26)
12.32%
I've had another experience (let us know in the thread!) (5)
2.37%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise